Worksheet 12 Solutions, Math 1B
Complex Numbers and Nonhomogeneous Linear Equations
Monday, April 16, 2012
1. Solve the differential equation or initial value problem using the method of undertermined coefficients.
(a) y 00 + 3y 0 + 2y = x2
Solution
We first find a solution to the homogeneous differential equation. the characteristic equation is
r2 + 3r + 2 = 0,
which has roots r = −1, −2 corresponding to solutions
y1 = e−t ,
y2 = e−2t .
We propose the particular solution yp = Ax2 + Bx + C, and we note that none of the linearly
independent parts x2 , x, and 1 are solutions to the homogeneous equation. Thus this proposed solution should be sufficient. We substitute the general proposed solution into the nonhomogeneous
equation to get
2A + 6Ax + 3B + 2Ax2 + 2Bx + 2C
= (2A)x2 + (6A + 2B)x + (2A + 3B + 2C) = x2 .
Equating the coefficients of corresponding linearly independent functions gives us the system of
equations


=1
2A
6A + 2B
=0,


2A + 3B + 2C = 0
which has solution A = 1/2, B = −3/2, and C = 7/4. Thus a particular solution to the differential
equation is
yp = x2 /2 − 3x/2 + 7/4,
and the general solution is
y = (x2 /2 − 3x/2 + 7/4) + C1 e−t + C2 e−2t .
(b) y 00 + 9y = e3x
Solution Idea
The homogeneous equation has solutions
y1 = cos(3x),
y2 = sin(3x),
and so the proposed form of the solution is
yp = Ae3x .
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(c) y 00 − 4y = ex cos x,
y(0) = 1,
y 0 (0) = 2
Solution Idea
The homogeneous equation has solutions
y1 = e2x ,
y2 = e−2x ,
so the proposed form of the solution is
yp = Aex cos x + Bex sin x.
2. Find all solutions to the equation x4 = 1.
Solution
Writing a complex number in polar form as z = reiθ , the equation is equivalent to
r4 ei(4θ) = 1.
Since both sides must have magnitude 1, we conclude that r = 1, and so we only need to find the
values of θ such that ei(4θ) = 1. Using Euler’s formula, this gives us
cos(4θ) + i sin(4θ) = 1,
and so we see that 4θ must be a multiple of 2π, or θ = kπ/2 for any integer k. But the values of
z corresponding to these values of θ repeat every 4 values, so we see that the only solutions to the
equation are

z = ei(0)
=0



z = ei(π/2) = i
.
z = ei(π)
= −1



z = ei(3π/2) = −i
3. Write a trial solution for the method of undertermined coefficients for the differential equation
y 00 + 3y 0 − 4y = (x3 + x)ex
Do not determine the coefficients.
Solution Sketch
The solutions to the homogeneous equation are
y1 = ex ,
y2 = e−4x .
We propose the form
(Ax3 + Bx2 + Cx + D)ex
for the trial solution, but notice that ex is a solution to the homogeneous equation. Thus we multiply
this form by x and propose
yp = (Ax4 + Bx3 + Cx2 + Dx)ex
for the trial solution. This shares no linearly independent parts with the homogeneous solution, so this
is the proper form for the trial solution.
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4. Use de Moivre’s Theorem with n = 3 to express cos 3θ and sin 3θ in terms of cos θ and sin θ.
Solution
De Moivre’s Theorem with r = 1 and n = 3 gives us that
cos 3θ + i sin 3θ = (cos θ + i sin θ)3 ,
and simplifying the right hand side, we have
cos 3θ + i sin 3θ = cos3 θ + 3i cos2 θ sin θ + 3i2 cos θ sin2 θ + i3 sin3 θ
= (cos3 θ − 3 cos θ sin2 θ) + i(3 cos2 θ sin θ − sin3 θ)
Equating real and imaginary parts gives us
cos 3θ = cos3 θ − 3 cos θ sin2 θ,
sin 3θ = 3 cos2 θ sin θ − sin3 θ.
5. Prove the following properties of complex numbers, where a line over a complex number indicates its
complex conjugate.
(a) z + w = z + w
Solution
Using rectangular coordinates, let z = a + ib, and let w = c + id. Then
z + w = (a + c) + i(b + d) = (a + c) − i(b + d) = (a − ib) + (c − id) = z + w.
(b) zw = zw
Solution
Using polar coordinates, let z = reiθ , and let w = seiφ . Then
zw = rsei(θ+φ) = rsei(−θ−φ) = rei(−θ) sei(−φ) = zw.
(c) z n = z n
Solution
Using polar coordinates, let z = reiθ . Then
n
z n = rn einθ = rn ei(−nθ) = rei(−θ) = z n
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